A solid sphere and a hollow sphere, both possessing the same mass and identical radii, are rolled down a rough inclined plane.

Step-by-step solution

Let M be the mass and R be the radius of both spheres. The moment of inertia of the solid sphere is I_S = 2 5 MR^2 and that of the hollow sphere is I_H = 2 3 MR^2 . For an object rolling down an inclined plane of height h , by conservation of mechanical energy: Mgh = 1 2 Mv^2 + 1 2 I ^2 Since v = R for pure rolling: Mgh = 1 2 Mv^2 (1 + I MR^2 ) v = 2gh 1 + I MR^2 Since I_S v_H . Thus, the solid sphere arrives at the bottom with a greater speed. The acceleration of a rolling body is a = g 1 + I MR^2 . Since I_S a_H